2

I have a skewed normal distribution for which I know the average, standard deviation, skewness & kurtosis (which is different from zero).

Given a number $X,$ how can estimate which percentile corresponds to that value? (I'm ok with getting an approximate value of this percentile.)

I used z-score tables in the past (before having skewed distributions), but they seem to apply only to non-skewed distributions.

Thanks for your help.

amWhy
  • 209,954
Nisalon
  • 151
  • Assuming you mean a skewed normal distribution (as defined at https://en.wikipedia.org/wiki/Skew_normal_distribution) you could use the method of moments to estimate the parameters of the skewed normal distribution and then estimate any desired function of that distribution. The problem is that you won't know how good your estimate of any percentile will be unless maybe you also have the sample size. – JimB Jan 07 '19 at 15:03
  • I have the sample size – Nisalon Jan 07 '19 at 16:16
  • If you have the sample size, you could sample from the estimated distribution to get some idea as to how good the parameter estimates might be. – JimB Jan 07 '19 at 16:59

1 Answers1

0

The Wiki page Skew normal distribution provides the information to estimate the parameters using the sample mean ($\bar{x}$), standard deviation ($s$), and skewness ($\hat{\gamma}$). The 3 parameters to be estimated are $\mu$, $\sigma$, and $\alpha$.

If $|\hat{\gamma}|<1$, then $\hat{\alpha}$ is found in two steps:

$$\delta =\sqrt{\frac{\pi \left| \hat{\gamma} \right| ^{2/3}}{2 \left(\left| \hat{\gamma} \right| ^{2/3}+\left(\frac{4-\pi }{2}\right)^{2/3}\right)}}$$

$$\hat{\alpha} = \text{sgn}(\hat{\gamma})\sqrt{\frac{\delta }{1-\delta ^2}}$$

Otherwise $\hat{\alpha}$ is the solution to

$$\hat{\gamma} =\frac{\sqrt{2} (4-\pi ) \hat{\alpha} ^3}{\left((\pi -2) \hat{\alpha} ^2+\pi \right)^{3/2}}$$

which needs to be performed numerically. Then $\hat{\mu}$ and $\hat{\sigma}$ are

$$\hat{\sigma} =\frac{s}{\sqrt{1-\frac{2 \hat{\alpha} ^2}{\pi \left(\hat{\alpha} ^2+1\right)}}}$$

$$\hat{\mu} =\bar{x}-\frac{\sqrt{\frac{2}{\pi }} \hat{\alpha} \hat{\sigma} }{\sqrt{\hat{\alpha} ^2+1}}$$

Now armed with the parameter estimates, then one can estimate the cumulative distribution function:

$$Pr(X \le x)=\Phi\left(\frac{x\, -\hat{\mu} }{\hat{\sigma} }\right)-2 T\left(\frac{x\, -\hat{\mu} }{\hat{\sigma} },\hat{\alpha} \right)$$

where $T$ is the Owen's T function: $T(x,a)=\frac{\int_0^a \frac{\exp \left(-\left(\left(t^2+1\right) x^2\right)\right)}{2 \left(t^2+1\right)} \, dt}{2 \pi }$ .

Here is an implementation using Mathematica:

Mathematica code

I know code should be given as text but in this case because it is unlikely that you have Mathematica (and it would look much messier as text), it should be instructive as to the process.

To estimate the percentage of the distribution no larger than a specified value you'll need to use the cumulative distribution function (CDF) described on the Wiki page. Using Mathematica for values of $X$ being 7.5 and 5.2:

Example of percentiles

If you have access to the statistical package R, then the sn package will perform these calculations.

JimB
  • 1,861